Problem: Let $P(x) = 0$ be the polynomial equation of least possible degree, with rational coefficients, having $\sqrt[3]{7} + \sqrt[3]{49}$ as a root.  Compute the product of all of the roots of $P(x) = 0.$
Explanation: We recognize the number $\sqrt[3]{7} + \sqrt[3]{49}$ from the difference-of-cubes factorization \[7 - 1 = \left(\sqrt[3]{7} - 1\right)\left(1 + \sqrt[3]{7} + \sqrt[3]{49}\right).\]Solving for $\sqrt[3]{7} + \sqrt[3]{49},$ we get \[\sqrt[3]{7} + \sqrt[3]{49} = \frac{7-1}{\sqrt[3]{7}-1} - 1 = \frac{6}{\sqrt[3]{7}-1} - 1.\]We can use this expression to build a polynomial which has $\sqrt[3]{7} + \sqrt[3]{49}$ as a root. First, note that $\sqrt[3]{7}$ is a root of $x^3 - 7 = 0.$ Then, $\sqrt[3]{7}-1$ is a root of $(x+1)^3 - 7 = 0,$ because $(\sqrt[3]{7}-1+1)^3 - 7 = (\sqrt[3]{7})^3 - 7 = 0.$ (You could also note that the graph of $y=(x+1)^3-7$ is a one-unit leftward shift of the graph of $y=x^3-7,$ so the roots of $(x+1)^3-7=0$ are one less than the roots of $x^3-7=0.$)

It follows that $\frac{6}{\sqrt[3]{7}-1}$ is a root of the equation \[\left(\frac{6}{x}+1\right)^3 - 7= 0,\]because when $x = \frac{6}{\sqrt[3]{7}-1},$ we have $\frac{6}{x} = \sqrt[3]{7}-1.$ We multiply both sides by $x^3$ to create the polynomial equation \[(6+x)^3 - 7x^3 = 0.\]Finally, replacing $x$ with $x+1$ like before, we see that $\frac{6}{\sqrt[3]{7}-1} - 1$ is a root of the equation \[(7+x)^3 - 7(x+1)^3 = 0.\]This equation is equivalent to \[x^3 - 21x - 56 = 0,\]so by Vieta's formulas, the product of the roots is $\boxed{56}.$